When a 20 g mass hangs attached to one end of a light spring of length 10 cm, the spring stretches by 2 cm. The mass is pulled down until the total length of the spring is 14 cm. The elastic energy, (in Joule) stored in the spring is :

Options

A. `4xx10^(-2)`
B. `4xx10^(-3)`
C. `8xx10^(-2)`
D. `8xx10^(-3)`

Answer

To solve the problem, we need to determine the elastic potential energy stored in the spring when it is stretched. We'll follow these steps:

### Step 1: Understand the given data
- Mass (m) = 20 g = 0.02 kg (convert grams to kilograms)
- Initial length of the spring = 10 cm
- Extension of the spring due to the mass = 2 cm
- Total length of the spring when pulled down = 14 cm
- Additional extension when pulled down = 14 cm - 10 cm = 4 cm

### Step 2: Calculate the spring constant (k)
Using Hooke's Law, which states that the force exerted by a spring is proportional to its extension:
\[ F = kx \]
Where:
- \( F \) is the weight of the mass (mg)
- \( k \) is the spring constant
- \( x \) is the extension of the spring

From the equilibrium condition:
\[ kx = mg \]
Substituting the values:
- \( m = 0.02 \, \text{kg} \)
- \( g = 9.81 \, \text{m/s}^2 \)
- \( x = 0.02 \, \text{m} \) (2 cm in meters)

Calculating the force:
\[ mg = 0.02 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 0.1962 \, \text{N} \]

Now substituting into Hooke's Law:
\[ k \times 0.02 \, \text{m} = 0.1962 \, \text{N} \]
\[ k = \frac{0.1962 \, \text{N}}{0.02 \, \text{m}} = 9.81 \, \text{N/m} \]

### Step 3: Calculate the elastic potential energy (EPE)
The elastic potential energy stored in the spring when it is stretched is given by the formula:
\[ EPE = \frac{1}{2} k x^2 \]
Where:
- \( x \) is the total extension from the original length of the spring.

In this case, the total extension when the spring is pulled down to 14 cm is:
\[ x = 14 \, \text{cm} - 10 \, \text{cm} = 4 \, \text{cm} = 0.04 \, \text{m} \]

Now substituting the values into the EPE formula:
\[ EPE = \frac{1}{2} \times 9.81 \, \text{N/m} \times (0.04 \, \text{m})^2 \]
\[ EPE = \frac{1}{2} \times 9.81 \times 0.0016 \]
\[ EPE = \frac{1}{2} \times 0.015696 \]
\[ EPE = 0.007848 \, \text{J} \]

### Step 4: Convert to appropriate units
To express the energy in a more standard form:
\[ EPE = 7.848 \times 10^{-3} \, \text{J} \]

### Final Answer
The elastic energy stored in the spring is approximately:
\[ EPE \approx 7.85 \times 10^{-3} \, \text{J} \]